Answer:
The pH of the solution is 12.61
Explanation:
Step 1: Data given
Concentration of hydrazoic acid solution = 1.200 M
Volume of the solution = 242.5 mL = 0.2425 L
Concentration of NaOH solution = 0.3400 M
The pKa of hydrazoic acid is 4.72
Step 2: The balanced equation
NaOH + HN3 → NaN3 + H2O
Step 3: Calculate moles hydrazoic acid
Moles hydrazoic acid (HN3) = concentration * volume
Moles hydrazoic acid = 1.200 M * 0.2425 L
Moles hydrazoic acid = 0.291 moles
Step 4: Calculate moles NaOH
Moles NaOH = 0.3400 M * 1.006 L
Moles NaOH = 0.342 moles
Step 5: Determine the limiting reactant
HN3 is the limiting reactant. It will completely be consumed (0.291 moles)
NaOH is in excess. There reacts 0.291 moles. There will remain 0.342 - 0.291 = 0.051 moles NaOH
Step 6: Calculate total volume
Total volume = 242.5 mL + 1006 mL = 1248.5 mL = 1.2485 L
Step 7: Calculate concentration of NaOH
concentration NaOH = 0.051 moles / 1.2485 L
concentration NaOH = 0.0408 M
Step 8: Calculate pOH
pOH = -log [OH-] = -log(0.0408)
pOH = 1.39
Step 9: Calculate pH
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 1.39
pH = 12.61
The pH of the solution is 12.61
